Entropy Changes & Processes
The Third Law of Thermodynamics
At absolute zero, or when T = 0 K, all energy of thermal motion has been quenched, and
all atoms or ions in a perfect crystalline lattice are in a perfect continuous array
Chapter 4 of Atkins: The Second Law: The Concepts
9 No spatial disorder
9 No thermal motion
9 Entropy is zero: if S = 0,
there is only one way of
arranging the molecules
Section 4.4 - 4.7
Third Law of Thermodynamics
Nernst Heat Theorem
Third- Law Entropies
Reaching Very Low Temperatures
Cannot actually reach absolute
zero - everything has some
internal energy!!
Helmholtz and Gibbs Energies
Helmholtz Energy
Maximum Work
Gibbs Energy
Maximum Non- Expansion Work
What good is the Third Law? It allows us to realize that as T approaches zero, the
absolute entropy tends towards zero. The effects of the third law are most keenly felt at
very low temperatures (not everyday stuff).
The third law also lets us define some entropies of substances relative to their perfect
crystals at 0 K. Useful book keeping device!
Standard Molar Gibbs Energies
Prof. Mueller
Chemistry 451 - Fall 2003
Lecture 12- 1
Nernst Heat Theorem
−402 J mol-1
= −1.09 J mol-1 K -1
369 K
Two entropies can be determined from measuring heat capacities
from T = 0 K to T = 369 K:
Sm (α ) = Sm (α ,0) + 37 J mol-1 K -1
Prof. Mueller
Chemistry 451 - Fall 2003
Lecture 12- 2
Third Law & The Nernst Heat Theorem
The entropy change accompanying a physical or chemical
transformation approaches zero as T approaches zero:
S → 0 as T → 0
Consider transition from orthorhombic sulfur S(α) to monoclinic
sulfur S(β) in the solid state. At the transition T (369 K):
∆ trsS = Sm (α ) − Sm (β ) =
Perfect continuous
array of atoms in
solid NaCl
Sm (β ) = Sm (β ,0) + 38 J mol-1 K -1
So at the transition temperature:
∆ trsS = Sm (α,0) − Sm (β,0) −1 J mol-1 K -1
If we decide to assign a value zero to entropies of elements in their
perfect crystalline form at T = 0, then all perfect crystalline compounds
have entropy S = 0 at T = 0
Third Law of Thermodynamics
If the entropy of every element in its most stable state at T = 0 is
taken as zero, then every substance has a positive entropy which
at T = 0 may become zero, which is also zero for all perfect
crystalline substances, including compounds
This does not mean that the entropy at T = 0 is really zero: it means that
all perfect crystalline substances have the same entropy at that
temperature (choosing the value S = 0 at this temperature is a
convenience, and as mentioned, leads to some very neat bookkeeping for
comparing relative entropies)
Sm (α,0) − Sm (β,0) = 0
Prof. Mueller
Chemistry 451 - Fall 2003
Lecture 12- 3
Prof. Mueller
Chemistry 451 - Fall 2003
Lecture 12- 4
1
Third Law Entropies
Reaching Low Temperatures
The choice S(0)=0 for perfect crystals is made from now on, and entropies reported
relative to this value are called Third Law Entropies (or just standard entropies)
A substance in its standard state at temperature T has a standard entropy which is
denoted as S∅(T)
Standard reaction entropies are defined as
∆ r S ⊕ = ∑ν J Sm⊕ (J )
J
and are weighted by stoichoimetric coefficients in the same way that enthalpies are
weighted
Example: Calculate ∆rS∅ of Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu (s) at 25oC
∆ r S ⊕ = Sm⊕ (Zn 2+ (aq))+ Sm⊕ (Cu(s))− Sm⊕ (Zn(s))− Sm⊕ (Cu 2+ (aq))
= [−112.1+ 33.15 − 41.63 + 99.6] J K−1 mol−1
−1
Chemistry 451 - Fall 2003
Lower temperatures (down to 20 nK) can be achieved by adiabatic
demagnetization - without a magnetic field, e- in paramagnetic
materials are oriented randomly; however, in the presence of a magnetic
field, the spin of the e- comes into play
There are more α spins (+1/2) than β spins (-1/2), and the entropy of the
sample is lowered (the spins are biased to point one way or the other)
Paramagnetic substances are cooled to 1 K with liquid He, and then the
application of a magnetic field lowers the energy of the unpaired electrons
isothermally (step A → B)
The spins are in a lower energy state, He is pumped off, the magnetic field
removed and thermal contact is broken (adiabatic demagnetization, step B
→ C). The sample is the same as at A, except with lower entropy and T.
−1
= −21.0 J K mol
Prof. Mueller
Temperatures below 4 K (the boiling point of He) can be achieved by
evaporating He through large radius pipes (down to 1 K) by JouleThomson expansion
Lecture 12- 5
Reaching Low Temperatures
Prof. Mueller
Chemistry 451 - Fall 2003
Lecture 12- 6
2001 Nobel Prize in Physics: BEC
Eric A. Cornell (NIST), Wolfgang Ketterle (MIT) & Carl E. Wieman (University of
Colorado)
"for the achievement of Bose- Einstein condensation in dilute gases of
alkali atoms, and for early fundamental studies of the properties of the
condensates".
In 1924 the Indian physicist Bose made important theoretical calculations
regarding light particles. Einstein predicted that if a gas of such atoms were
cooled to a very low temperature all the atoms would suddenly gather in the
lowest possible energy state. Seventy years were to pass before these
Nobel Laureates, in 1995, succeeded in achieving this extreme state of matter.
Cornell and Wieman then produced a pure condensate of about 2 000 rubidium
atoms at 20 nK (nanokelvin), i. e. 0.000 000 02 degrees above absolute zero.
Independently of the work of Cornell and Wieman, Ketterle performed
corresponding experiments with sodium atoms.
Please visit: http:// www. nobel. se to see more about this year’s prizes, esp. the
prize for the development of MRI.
Prof. Mueller
Chemistry 451 - Fall 2003
Lecture 12- 7
Prof. Mueller
Chemistry 451 - Fall 2003
Lecture 12- 8
2
Helmholtz and Gibbs Energies
Helmholtz and Gibbs Energies, 2
We will now focus almost exclusively on entropy changes within the system
(treating the surroundings and total entropy is trivial)
Consider a system in thermal equilibrium with surroundings, at temperature T. For
a change in the system with transfer of heat, the Clausius inequality is:
Heat transfer at constant V
dU
≥0
dS −
T
TdS ≥ dU
dS −
dq
≥0
T
Heat transfer at constant p
(no non-expansion work)
dH
≥0
T
TdS ≥ dH
dS −
In both cases, criteria for spontaneous change are expressed solely in terms of state
functions. Below, subscripts indicate constant properties:
Constant dH = 0 or dS = 0:
Constant dU = 0 or dS = 0:
dSU ,V ≥ 0
Prof. Mueller
dS H , p ≥ 0
dU S,V ≤ 0
dH S, p ≤ 0
Chemistry 451 - Fall 2003
Lecture 12- 9
To interpret these inequalities, consider the constant V cases:
dSU ,V ≥ 0
Inequalities on previous slide reformed as dU –TdS ≤ 0 and dH - TdS ≤ 0,
we can write them as two new thermodynamic quantities:
Helmholtz Energy,A
Gibbs Energy,G
A = U − TS
dAT ,V ≤ 0
Prof. Mueller
Chemistry 451 - Fall 2003
G = H − TS
dG = dH − TdS
dGT , p ≤ 0
Lecture 12- 10
Maximum Work
Change in a system at constant temperature and volume is spontaneous if
dAT,V ≤ 0 (change under these conditions corresponds to decrease in A)
A = U − TS
dA = dU − TdS
Helmholtz Energy is useful also for indicating the maximum amount of
work accompanying a process (A called maximum work function)
dwmax = dA
How do we prove this? Combine the Clausius inequality, dS ≥ dq/T in the
form TdS ≥ dq with the first law U = q +w
dAT ,V ≤ 0
dU ≤ T dS + dw
Equilibrium condition for complete reversibility, is when dAT,V = 0
The above expressions can be interpreted as follows:
A negative dA is favored by negative dU and positive TdS
Does system move to lower A due to tendency to move towards states of
lower internal energy and higher entropy? No, tendency towards lower A is
solely from a move towards states of greater overall entropy
Systems change spontaneously if the total entropy of system and
surroundings increase, not because system tends towards lower internal
energy: dS is change in system entropy,-dU/T is the entropy change of
surroundings (when V constant), and their total tends to a maximum
Chemistry 451 - Fall 2003
at constant T
dA = dU − TdS
Helmholtz Energy
Prof. Mueller
dU S,V ≤ 0
(1) The first inequality says that for a system at constant volume and
constant internal energy (e.g., isothermal system), entropy increases in a
spontaneous change
(2) The second inequality says if S and V are constant, then internal
energy must decrease with spontaneous change - system doesn’t
spontaneously go to lower energy; rather, if ∆Ssys= 0, then ∆Ssur > 0, if
energy flows out of system as heat, and system energy decreases.
Lecture 12- 11
dU is smaller than the RHS b/c we replace dq byTdS (which is in general
larger), and rearranging:
dw ≥ dU – T dS
The most negative value of dw, and therefore the maximum energy
obtained from the system as work is
dwmax = dU – T dS
which only applies on a reversible path (hence the equality in the
equation), and dwmax= dA at constant T
Prof. Mueller
Chemistry 451 - Fall 2003
Lecture 12- 12
3
Maximum Work, 2
Maximum Work, 3
For macroscopic measurable change:
wmax = ∆A
where ∆A = ∆U – T ∆S
Depending upon the sign of T∆S, not all of ∆U is available for work
Helmholtz Free Energy, ∆A is the part of change in internal energy that
we are free to use for work
Molecular Intrepretation: A is total internal energy U, less a chaotically
stored contribution, T S
If T ∆S < 0, the RHS is not as negative as ∆U itself, so maximum work is
less than ∆U. For change to be spontaneous, some energy must escape as
heat to generate enough entropy in surroundings to overcome reduction in
entropy in system (∆STOT>0)
If T ∆S > 0, the RHS is more negative than ∆U itself, so maximum work is
more than ∆U. System is not isolated, and heat can flow in as work is
done. Some reduction in Ssur occurs, yet overall ∆STOT>0
Prof. Mueller
Chemistry 451 - Fall 2003
Lecture 12- 13
Prof. Mueller
Calculating Maximum Work
When 1.000 mol C6H12O6 is burned at 25oC:
C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O (l)
calorimeter measurements give ∆rU o = –2808 kJ mol-1 and ∆rS o = + 182.4 J K
mol-1 at 25oC. How much of the energy is extracted as (a) heat at constant
pressure and (b) work?
(a) ∆ng=0, so ∆rU o = ∆rH o= –2808 kJ mol-1. So at constant pressure, energy
available as heat is q = 2808 kJ mol-1
(b) At T=298 K, we can calculate ∆rAo as
∆ rAo = ∆rU o – T ∆rSo = –2862 kJ mol-1
So what this means is that burning glucose in oxygen can be used to produce a
maximum of 286s2 kJ mol-1 of work
The maximum energy available for work is greater than the change in internal
energy of the system due to the positive entropy change in the system
(generating small molecules from one big molecule) – the system therefore
draws energy from the surroundings (reducing their entropy) for doing work
Prof. Mueller
Chemistry 451 - Fall 2003
Lecture 12- 15
Chemistry 451 - Fall 2003
Lecture 12- 14
Gibbs Energy
The Gibbs energy, or free energy, is more commonly used in chemistry
because we are interested mainly in constant pressure changes as
opposed to constant volume changes
G = H − TS
dG = dH − TdS
dGT , p ≤ 0
The inequality dGT,p ≤ 0 tells us that at constant temperature and
pressure, chemical reactions are spontaneous in the direction of
decreasing Gibbs energy
9 If G decreases as a reaction occurs, the conversion of reactants into
products is spontaneous
9 If G increases during a reaction, the reverse reaction is spontaneous
The existence of spontaneous endothermic reactions can be
explained with G: H increases (spontaneously higher enthalpy), so dH > 0.
The dG of this spontaneous reaction is < 0: the entropy increase must be
high enough that TdS is larger and positive, and outweighs dH
Prof. Mueller
Chemistry 451 - Fall 2003
Lecture 12- 16
4
Maximum Non-Expansion Work
Maximum non-expansion work, wadd, is given by change in Gibbs energy
dwadd,max = dG, wadd,max = ∆G
Since H = U + pV for a general change, then
dH‘ = dq + dw + d(pV)
Remembering that G = H-TS, then if the change is reversible, dw = dwrev
and dq = dqrev = T dS. Thus for a reversible and isothermal process:
dG‘ =TdS + dwrev+ d(pV) - TdS‘ = dwrev + d(pV)
Work consists of expansion work, dwrev=-p dV, and some other kind of
work (e.g., electrical work pushing electrons through a circuit, work raising
a column of fluid, etc.). The latter work is non-expansion work, dwadd
dG‘ = (-pdV + dwadd,rev) + pdV + Vdp
= dwadd,rev + Vdp
Calculating Maximum Non-Expansion Work
How much energy is available for sustaining muscular and nervous system
activity from combustion of 1.00 mol of glucose under standard conditions
at 37oC (temperature of blood)? ∆rH o = -2808 kJ mol-1
and ∆rS o = +182.4 J K-1mol-1
C6H12O6(s) + 6O2(g) → 6CO2(g) +6H2O (l)
The non-expansion work available from the combustion can be calculated
from the standard Gibbs energy, ∆rG o = ∆rH o –T∆rS o
∆rG o = -2808 kJ mol-1 - (310 K) x (182.4 J K-1mol-1) =-2865 kJ mol-1
so, wadd,max =-2865 kJ mol-1 for combustion of 1 mole of glucose molecules
in blood, and the reaction can be used to do up to 2865 kJ mol-1 of nonexpansion work !
And, if work occurs at constant p, then dG = dwadd,rev; because the
process is reversible, work has its maximum value here !
Prof. Mueller
Chemistry 451 - Fall 2003
Lecture 12- 17
Prof. Mueller
Chemistry 451 - Fall 2003
Lecture 12- 18
Standard Molar Gibbs Energies
The non-expansion work available from the combustion can be calculated
from the standard Gibbs energy, ∆rG o
∆r G o = ∆r H o – T ∆r S o
The standard Gibbs energy of formation, ∆fG o, is the standard
reaction Gibbs energy for the formation of a compound from its elements
in their reference states – ∆rG o can be expressed in terms of ∆fG o :
∆ rG ⊕ = ∑ν J ∆ f G ⊕ (J )
J
∆fG o = 0 for elements in their reference states, since their formation is a
null reaction (see table 4.4 Atkins 7th Edition, or CRC Handbook)
Calorimetry (for ∆H directly and ∆S via heat capacities) is one way of
determining Gibbs energies - however, they can be determined as well
from reaction equilibrium constants, electrochemical measurements and
spectroscopic measurements…
Prof. Mueller
Chemistry 451 - Fall 2003
Lecture 12- 19
5